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The Elements (Longer Version)
Ian Beardsley!
(University of Oregon, Department of Physics, 2022)!
!
Genesis Project California 2022!
⋅
⋅
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We see that the periodic tables is cyclical over 18 groups of periodicity of 18 that is the
interplay of 2 and 3 through a factor of six. We see this is the basis for many ancient
music traditions over the globe, and not only predicts the radius of a proton in terms of
carbon the core element of life (element 6) but explains the ancient Greek formulation of
the elements as Earth, Air, Water, and Fire with modern chemistry.!
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Section 1.0 Overview
Throughout ancient human history, measures of six have been used for the passage of time.
There are the romantic guzals of India, the bajans and kirtans. In the Middle East six is played
as three, two times. In Persia it is played as two, three times. In the flamenco of Spain it is
played as 12: two three’s plus three two’s.!
In Ancient Greece they classified the elements as four kinds: Earth, Air, Water, and Fire. In
modern terms we could call this the phases of matter which is solid is earth, air is a gas, water
is a liquid, and fire is energy (the emission of photons from an excited atom). !
However, today we can surmise the elements are discrete units (protons) with electrons orbiting
them. The number of protons determines the type of element and it has a number of electrons
equal to that number so that it is electrically neutral. For instance hydrogen is one proton, with
one electron. Carbon is six protons with six electrons. We find their properties are periodic over
18 groups, the last group being the inert or noble gases. When one element combines with
another in the case of a metal with a non-metal, the metal loses electrons to become positive
and the non-metal gains electrons to become negative. The period of 18 groups determines
how many electrons are gained or lost because the non-metal wants to have noble gas
electron configuration. So it is in oxygen in group16 that it gains 18-16=3 electrons becoming
. If it is aluminum in group 13, then it becomes . This determines how aluminum
combines with oxygen to become aluminum oxide because it must be neutral. Since Al has a
charge of positive three it needs 3(2)=6 (2 atoms) to have a positive charge of 6. And since
oxygen has a negative charge of minus 2 it needs 2(3)=6 (3 atoms) to have a negative charge of
minus six. Thus aluminum combines with oxygen as two aluminum atoms to three oxygen
atoms, and we write its chemical formula as .!
Thus we see since the periodic table is periodic over 18 groups it is deeply connected to the
six that occurs in the guzals of India, the sixes of the Middle Eastern and Persian music, and
the six of Spain’s flamenco. Indeed the key to playing in this meter is in understanding the
relationship between 3 and 2 which defines the syncopated nature of the these styles, which is
the reconciliation between these two numbers. It occurs by way of the factor of six over a
period of 18. Eighteen as I said is the periodicity of the periodic table of the elements, and I
would like to say now that carbon (C) is the sixth element and the core element of life because
it becomes when it combines with other elements, which is for the skeleton of life the
hydrocarbons by combing with hydrogen , which is one proton. I have found that carbon (C)
predicts the radius of this proton. Let us look at this now in the following section, section 2.0.
We will look after that at how the this interplay of 2 and 3 reconciles the Ancient Greek
categorization of the Elements — Earth, Air, Wind, Fire — with the modern formulations of
elemental compounds in terms of atoms.!
2.0 Proton-Seconds and Proton Radius
We need to have one equation that brings in constants of quantum mechanics and of
relativity. Like of QM and of gravity. I find I can do this with inertia.!
In order to present the elements as mathematical structures we need to explain the matter from
which they are made as mathematical constructs. We need a theory for Inertia. I had found
(Beardsley Essays In Cosmic Archaeology. 2021) where I suggested the idea of proton
seconds, that is six proton-seconds, which is carbon the core element of biological life if we
can figure out a reason to divide out the seconds. I found!
O
2−
Al
3+
Al
2
O
3
C
4−
H
+
h
G
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Equation 2.1 !
Where h is Planck’s constant 6.62607E—34 Js, is the radius of a proton 0.833E-15m, G is
the universal constant of gravitation 6.67408E-11 (Nm2)/(kg2), and c is the speed of light
299,792,459 m/s. And is t=1 second. is the Sommerfeld constant (or fine structure
constant) is 1/137. The mass of a proton is .!
Since plank’s constant h is a measure of energy over time where space and time are
concerned it must play a role. Of course the radius of a proton plays a role since squared and
multiplied by it is the surface area of our proton embedded in space. The gravitational
constant is force produced per kilogram over a distance, thus it is a measure of how the
surrounding space has an effect on the proton giving it inertia. The speed of light c has to play
a role because it is the velocity at which event are separated through time. The mass of a
proton has to play a role because it is a measurement of inertia itself. And alas the fine
structure constant described the degree to which these factors have an effect. We see the
inertia then in equation 3.1 is six protons over 1 second, by dimensional analysis. !
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed
of light squared: !
!
Matter is that which has inertia. This means it resists change in position with a force applied to
it. The more of it, the more it resists a force. We understand this from experience, but what is
matter that it has inertia?!
I would like to answer this by considering matter in one of its simplest manifestations, the
proton, a small sphere with a mass of 1.6726E-27 kg. This is a measure of its inertia.!
I would like to suggest that matter, often a collection of these protons, is the three dimensional
cross-section of a four dimensional hypersphere.!
The way to visualize this is to take space as a two-dimensional plane and the proton as a two
dimensional cross-section of a sphere, which would be a circle.!
In this analogy we are suggesting a proton is a three dimensional bubble embedded in a two
dimensional plane. As such there has to be a normal vector holding the higher dimensional
sphere in a lower dimensional space. Thus if we apply a force to to the cross-section of the
sphere in the plane there should be a force countering it proportional to the normal holding it in
a lower dimensional universe. This counter force would be experienced as inertia. It may even
induce in it an electric field, and we can see how it may do the same equal but opposite for the
electron.#
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6protons
r
p
t
1
α
m
p
= 1.67262E − 27kg
4π
α
2
=
U
e
m
e
c
2
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#
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Since we have!
!
Is a number of protons!
!
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided
by the mass of a proton. But these masses can be considered to cancel and leave pure
number.!
We make a program that looks for close to whole number solutions so we can create a table of
values for problem solving. I set it at decimal part equal to 0.25. You can choose how may
values for t you want to try, and by what to increment them. Here are the results for
incrementing by 0.25 seconds then 0.05 seconds. Constant to all of this is hydrogen and
carbon. The smaller integer value of seconds gives carbon (6 protons at 1 second) and the
largest integer value of seconds gives hydrogen (1 proton at six seconds) and outside of that
for the other integer values of protons you get are at t>0 and t<1. Equation 1 really has some
interesting properties. Here are two runs of the program( decpart is just me verifying that my
boolean test was working right to sort out whole number solutions):!
By what value would you like to increment?: 0.25!
How many values would you like to calculate for t in equation 1 (no more than 100?): 100!
24.1199 protons 0.250000 seconds 0.119904 decpart !
12.0600 protons 0.500000 seconds 0.059952 decpart !
8.0400 protons 0.750000 seconds 0.039968 decpart !
6.0300 protons 1.000000 seconds 0.029976 decpart !
4.0200 protons 1.500000 seconds 0.019984 decpart !
3.0150 protons 2.000000 seconds 0.014988 decpart !
2.1927 protons 2.750000 seconds 0.192718 decpart !
2.0100 protons 3.000000 seconds 0.009992 decpart !
1.2060 protons 5.000000 seconds 0.205995 decpart !
1.1486 protons 5.250000 seconds 0.148567 decpart !
1.0964 protons 5.500000 seconds 0.096359 decpart !
1.0487 protons 5.750000 seconds 0.048691 decpart !
1.0050 protons 6.000000 seconds 0.004996 decpart !
0.2487 protons 24.250000 seconds 0.248659 decpart !
0.2461 protons 24.500000 seconds 0.246121 decpart !
0.2436 protons 24.750000 seconds 0.243635 decpart !
By what value would you like to increment?: 0.05!
How many values would you like to calculate for t in equation 1 (no more than 100?): 100!
40.1998 protons 0.150000 seconds 0.199837 decpart !
30.1499 protons 0.200000 seconds 0.149879 decpart !
24.1199 protons 0.250000 seconds 0.119904 decpart !
20.0999 protons 0.300000 seconds 0.099918 decpart !
17.2285 protons 0.350000 seconds 0.228500 decpart !
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
1
α
2
m
p
h4πr
2
p
Gc
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15.0749 protons 0.400000 seconds 0.074938 decpart !
12.0599 protons 0.500000 seconds 0.059950 decpart !
10.0500 protons 0.600000 seconds 0.049958 decpart !
8.0400 protons 0.750000 seconds 0.039966 decpart !
7.0941 protons 0.850000 seconds 0.094088 decpart !
6.0300 protons 1.000000 seconds 0.029975 decpart !
5.2435 protons 1.150000 seconds 0.243457 decpart !
5.0250 protons 1.200000 seconds 0.024980 decpart !
4.1586 protons 1.450000 seconds 0.158605 decpart !
4.0200 protons 1.500000 seconds 0.019985 decpart !
3.1737 protons 1.899999 seconds 0.173673 decpart !
3.0923 protons 1.949999 seconds 0.092296 decpart !
3.0150 protons 1.999999 seconds 0.014989 decpart !
2.2333 protons 2.699999 seconds 0.233325 decpart !
2.1927 protons 2.749999 seconds 0.192719 decpart !
2.1536 protons 2.799999 seconds 0.153564 decpart !
2.1158 protons 2.849998 seconds 0.115782 decpart !
2.0793 protons 2.899998 seconds 0.079303 decpart !
2.0441 protons 2.949998 seconds 0.044061 decpart !
2.0100 protons 2.999998 seconds 0.009993 decpart !
1.2433 protons 4.850000 seconds 0.243294 decpart !
1.2306 protons 4.900001 seconds 0.230607 decpart !
1.2182 protons 4.950001 seconds 0.218177 decpart !
Here is the code for the program:!
#include <stdio.h>!
#include <math.h>!
int main(int argc, const char * argv[]) {!
!
int n;!
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;!
!
do!
{!
printf("By what value would you like to increment?: ");!
scanf("%f", &increment);!
printf("How many values would you like to calculate for t in equation 1 (no more than 100?):
");!
scanf("%i", &n);!
}!
while (n>=101);!
{!
for (int i=0; i<n;i++)!
{!
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));!
!
int intpart=(int)protons[i];!
float decpart=protons[i]-intpart;!
t=t+increment;!
if (decpart<0.25)!
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{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);!
}}}}!
A very interesting thing here is looking at the values generated by the program, the smallest
integer value, 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest
of the elements heavier than carbon are formed by fractional seconds. These are the
hydrocarbons the backbones biological chemistry.!
Equation 2.2 !
!
!
!
Thus we have six protons (carbon) is one second. We see 1 second predicts the radius of a
proton,…!
In that we have!
!
And the periodic table of the elements is cyclical with 18 groups and!
!
Then perhaps we are supposed to write!
Equation 2.3. !
In fact, what if the 3 is supposed to be pi, then!
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352seconds
1
α
2
m
p
h4πr
2
p
Gc
= (6proton s)(1secon d )
1
α
2
m
p
h4πr
2
p
Gc
= (1proton)(6secon ds)
α
2
=
U
e
m
e
c
2
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
6 =
1
α
2
m
p
h4πr
2
p
Gc
3
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
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!
Then we would say that!
k=18/pi=5.7229577951!
The parameter in our constant with the most uncertainty is the radius of a proton . If the 3 is
supposed to be pi, then the radius of a proton becomes:!
Equation 2.4 !
Which gives!
!
About 95% raw most recent value measured. But, if !
!
Is supposed to be 6 and it is supposed to be multiplied by three to give 18 even which we
need for chemistry so we have 18 protons in the last group of the periodic table which is
important because we need argon with 18 protons for predicting valence numbers of elements
in terms of their need to attain noble gas electron configuration. Then we get!
!
This is in very close agreement with the most recent value measured which is!
!
Thus we see carbon (we will call the six-factor crux of the periodic table) determines the the
founding element of the periodic table, hydrogen with its one proton. In the next section,
section 3.0, we will now address how this same interplay reconciles the ancient Greek
classification of the elements as Earth, Air, Water, and Fire through the modern geometric
theory of atomic structure.#
π
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
r
p
r
p
= kα
2
m
p
Gc
4πh
r
p
= 8.790587E − 16m
1
α
2
m
p
h4πr
2
p
Gc
r
p
= 8.288587E − 16m = 0.829f m
r
p
= 0.833 + / − 0.014
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Section 3.0 Earth, Air, Water, Fire
In chemistry there is a simple theory for that explains the shapes of molecules called (VSEPR,
Valence Shell Electron Pair Repulsion). It states molecules with a central atom of say carbon,
which combine with four atoms of chlorine makes a pyramid. Not the pyramid of say Egypt,
which has a square base, but the tetrahedron, the four-faced pyramid with all faces a triangle.
We see the the bond angles will be 109.5 degrees because this is the result if we inscribe such
a tetrahedron in a cube, and calculate the internal angles by simple trigonometry as you will
conclude from the illustration below:!
!
If the the pyramid has a free pair of electrons for bonding it is called trigonal pyramidal and has
a bond angle of a little less than 109.5 degrees because the free electron push the other
elements bonded with the center atom, a little closer together.!
I have taken water, of course, as Ancient Greek water, Earth as silicon dioxide ( ) because
this is the primary component of sand, fire as methane ( ) because burned in oxygen it
emits light (is fire). And Ancient Greek air is nitrogen gas 78% and oxygen gas 21% . As
it turns out all of these have pyramidal geometries. We indeed want to look at ammonia ( )
as well because it is believed the primordial Earth was ammonia, methane, and water vapor
SiO
2
CH
4
N
2
O
2
NH
3
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which have been shown by experiment when combined under energy such as electricity (like
lightening) by Miller and Urey, form amino acids, the building blocks of life.!
(Referring to the illustration on the next page) We have that ammonia is trigonal pyramidal
because three it has hydrogens for the base of a pyramid and its apex is a free electron pair
giving it a bond angle a little less than the 109.5 degrees. It is by experiment 107 degrees.
However, methane is tetrahedral because three hydrogen atoms form its base but the
apex is another hydrogen atom meaning it does not have a free electron pair and thus has
exactly the 109.5 degree bond angle. Silicon having only two oxygens is triangular, or bent. But
when it connects with other SiO2 molecules it shares oxygen molecules with its neighbors
making it covalently . Thus with three oxygen O’s at its base, it has one oxygen O at its
top making it tetrahedral and giving it a bond angle of exactly 109.5 degrees. Water, similarily is
bent or triangular like SiO2 but it has a base of 3H and a lone electron pair making it have a
bond angle a little less than 109.5 degrees. It is at 104.5 degrees. This is all clear in illustration
on the next page…#
NH
3
CH
4
SiO
4
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#
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We see this is characteristic of the same interplay of the equation that gave the radius of a
proton in terms of carbon (element 6) that characterizes the 18 group periodicity of the Periodic
Table of the Elements, equation 2.3:!
!
And the music styles of India, The Middle East, India, Persia, and we should include Africa in
general. This because in India, 6 is often played counted an elongated one-two, that is the first
beat of the first 3 is counted one, and the first beat of the second group of 3 is counted 2 as
the turn around point. In Persia six is played 1 2 1 2 1 2. That is as three groups of 2. However
both play with their meters, India playing two’s over three, and Persia playing three’s over 2.
Flamenco often plays 6, three times, to make 18 in groups of 3, but finish with six in groups of
two. The result being three times six is 18 plus another six is 24 to make two cycles of 12.!
Thus we see the dynamics of these music forms are based on six in a way that reconciles 2
with 3. Eighteen is the number that reconciles 2 with 3, in the periodic table by way of the
factor of 6 (carbon) the core element of life. That is:!
3+3+3=9!
2+2+2=6!
!
!
!
We can look at this geometrically. We inscribe a triangle in a square because the interplay
between 2 and 3 would be the square and triangle, the square with two sides the same and the
equilateral triangle with 3 sides the same. The center of a polygon to its corner is its radius r.
We have the following illustration:!
3
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
3 ⋅ 6 = 18
2 ⋅ 9 = 18
2 ⋅ 3 = 6
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Rhythmically the two’s are shorter in duration than the three’s
because there are 3 of them in the same time there are only two
of the three’s. Thus they are the radii of the triangle and the
three’s are the sides.!
!
We can compute the angles and and find they are 48.59
degrees and 41.41 degrees respectively. (See illustration next
page)!
!
And The angles and (Illustration next page). Which are 30
degrees and 60 degrees respectively.!
!
Everything comes together in 6-fold symmetry which is
carbon (six protons) where each side equals its radius
made up of 60 degree triangles which make angles with
the apothem of 30 degrees each (drawing in the altitude
which bisects the 60 degree angles).#
α
β
ϕ
θ
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!
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Section 4.0 The Unifying Equation
We can bring all of this together with one equation by considering the square and equilateral
triangle. If we draw in the diagonal of the square we make an angle of 45 degrees. This means
its hypotenuse is . The cosine of 45 degrees is . We have!
Expression 4.1. !
If we draw in the altitude of the the equilateral triangles formed by triangles made by the
equilateral triangles radii, we have its radius is . This means!
Expression 4.2. !
Theorem 4.1. If we add the half-square to the square it is the counterpart to the half of the
equilateral triangle making a rectangle (See illustration next page). This half triangle if we draw
in the circle with the radius of its shorter side is the golden ratio . The golden ratio is:!
Expression 4.3. !
This results in the following unifying equation:!
Equation 4.1. !
n=4,5,6!
2
2/2
2 cos
π
4
1/ 3
2 cos
π
6
Φ
2 cos
π
5
f (n) = 2cos
π
n
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#
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Section 5.0 The Wave
Since we have in section 2.0 assigned time values to numbers of protons, and thus elements,
from equation 1.1!
!
And written a program that determines them:!
24.1199 protons 0.250000 seconds 0.119904 decpart !
12.0600 protons 0.500000 seconds 0.059952 decpart !
8.0400 protons 0.750000 seconds 0.039968 decpart !
6.0300 protons 1.000000 seconds 0.029976 decpart !
4.0200 protons 1.500000 seconds 0.019984 decpart !
3.0150 protons 2.000000 seconds 0.014988 decpart !
2.1927 protons 2.750000 seconds 0.192718 decpart !
2.0100 protons 3.000000 seconds 0.009992 decpart !
1.2060 protons 5.000000 seconds 0.205995 decpart !
1.1486 protons 5.250000 seconds 0.148567 decpart !
1.0964 protons 5.500000 seconds 0.096359 decpart !
1.0487 protons 5.750000 seconds 0.048691 decpart !
1.0050 protons 6.000000 seconds 0.004996 decpart !
0.2487 protons 24.250000 seconds 0.248659 decpart !
0.2461 protons 24.500000 seconds 0.246121 decpart !
0.2436 protons 24.750000 seconds 0.243635 decpart !
And we have angles associated with the elements from their geometry, we have a the equation
of a wave:!
Equation 5.1. !
If we say is the amplitude and t=1 second is carbon, then trying
degrees from our triangles and squares:!
!
!
!
And for hydrogen with a value of 6 seconds is:!
!
!
!
This is interesting to me because hydrogen being the basis of the periodic table is what it
should be, oscillating between -1 and 1 with the triangle and is 0 for the square. Carbon being
the 6 factor crux of the periodic table and core element of life chemistry is the dynamic ratios
involving the square root of three, and one half.#
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6proton s
A = A
0
cosωt
A
0
= 1
ω = 30,60,45
A(60) = cos(60 ⋅ 1) = 0.5
A(30) = cos(30 ⋅ 1) = 3/2
A(45) = cos(45 ⋅ 1) = 2/2
A(60) = cos(60 ⋅ 6) = 1
A(30) = cos(30 ⋅ 6) = − 1
A(45) = cos(45 ⋅ 6) = 0
of 19 45
Now that we have established a mathematical basis for the elements, the protons of
which they are made, and some of the more basic ensuing compounds, combinations of
elements such water (hydrogen and oxygen) ammonia (nitrogen and hydrogen) methane
(carbon and hydrogen) air (Nitrogen and Oxygen) the hydrocarbons (combination of
hydrogen and oxygen in long chains, the skeleton of life chemistry, or organic chemistry).
Indeed the biological elements have carbon at their center, geometrical and conceptually.
But at the center of artificial intelligence we have silicon, doped with boron and
phosphorus or gallium and arsenic to make semiconductors, the basis of the circuitry
used to make computing machines. We will see now that they are mathematical
constructs as well that abide by the geometry of three dimensional space as seen in
vector calculus.
Section 6.0 Artificial And Biological
We wish to establish a theorem from all that has been said here.!
Theorem 1.0 The Periodic Table of the Elements employs six fold symmetry. The periods of the
periodic table are periodic over 18 groups, which is 3 times 6, the three being the components
of , the number of values needed to specify a point in such a space.!
Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the elements. As
you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and germanium are in
group 14 meaning they have 4 valence electrons and want 4 for more to attain noble gas electron
configuration. If we dope Si with B from group 13 it gets three of the four electrons and thus has a
deficiency becoming positive type silicon and thus conducts. If we dope the Si with P from group 15 it
has an extra electron and thus conducts as well. If we join the two types of silicon we have a
semiconductor for making diodes and transistors from which we can make logic circuits for AI.
As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the right of Si
but doping agent B is not directly to the left, aluminum Al is. This becomes important. I call (As-Ga) the
differential across Ge, and (P-Al) the differential across Si and call Al a dummy in the differential because
boron B is actually used to make positive type silicon.
That the AI elements make a three by three matrix they can be organized with the letter E with subscripts
that tell what element it is and it properties, I have done this:
ℝ
3
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
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Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the first row
and has 4 valence electrons making it carbon (C). I believe that the AI elements can be organized in a 3 by
3 matrix makes them pivotal to structure in the Universe because we live in three dimensional space so
the mechanics of the realm we experience are described by such a matrix, for example the cross product.
Hence this paper where I show AI and biological life are mathematical constructs and described in terms
of one another.
We see, if we include the two biological elements in the matrix (E14) and and (E15) which are carbon and
nitrogen respectively, there is every reason to proceed with this reasoning if the idea is to show not only
are the AI elements and biological elements mathematical constructs, they are described in terms of one
another. We see this because the first row is ( B, C, N) and these happen to be the only elements that are
not core AI elements in the matrix, except boron (B) which is out of place, and aluminum (Al) as we will
see if a dummy representative makes for a mathematical construct, the harmonic mean. Which means we
have proved our case because the first row if we take the cross product between the second and third rows
are, its respective unit vectors for the components meaning they describe them:
And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological elements C
and N being the unit vectors are multiplied by the AI elements, meaning they describe them! But we have
to ask; Why does the first row have boron in it which is not a core biological element, but is a core AI
element? The answer is that boron is the one AI element that is out of place, that is, aluminum is in its
place. But we see this has a dynamic function.
A = (Al, Si, P)
B = (G a, G e, A s)
A ×
B =
B
C
N
Al Si P
G a G e A s
= (Si ⋅ A s − P ⋅ Ge)
B + (P ⋅ G a − Al ⋅ As)
C + (Al ⋅ Ge − Si ⋅ G a)
N
A ×
B = − 145
B + 138
C + 1.3924
N
A = 26.98
2
+ 28.09
2
+ 30.97
2
= 50g /m ol
B = 69.72
2
+ 72.64
2
+ 74.92
2
= 126g /m ol
A ⋅
B = A Bcosθ
cosθ =
6241
6300
= 0.99
θ = 8
∘
A ×
B = A Bsi nθ = (50)(126)si n8
∘
= 877.79
877.79 = 29.6g /m ol ≈ Si = 28.09g /m ol
of 21 45
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon (Si) and
germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic (As) have an
asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C) and as such have 4
valence electrons. Thus to have positive type silicon and germanium, they need doping agents from group
13 (three valence electrons) like boron and gallium, and to have negative type silicon and germanium they
need doping agents from group 15 like phosphorus and arsenic. But where gallium and arsenic are in the
same period as germanium, boron is in a different period than silicon (period 2) while phosphorus is not
(period 3). Thus aluminum (Al) is in boron’s place. This results in an interesting equation.
Equation 6.1
The differential across germanium crossed with silicon plus the differential across silicon crossed with
germanium normalized by the product between silicon and germanium is equal to the boron divided by
the average between the germanium and the silicon. The equation has nearly 100% accuracy (note: using
an older value for Ge here, it is now 72.64 but that makes the equation have a higher accuracy):
Due to an asymmetry in the periodic table of the elements due to boron we have the harmonic
mean between the semiconductor elements (by molar mass):
Equation 6.2.
This is Stokes Theorem if we approximate the harmonic mean with the arithmetic mean:
We can make this into two integrals:
Si(A s − G a) + Ge(P − Al )
SiG e
=
2B
Ge + Si
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(As − G a) +
Ge
B
(P − Al ) =
2SiG e
Si + G e
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
∫
1
0
∫
1
0
[
Si
B
(As − G a) +
Ge
B
(P − Al )
]
d x d y ≈
1
Ge − Si
∫
Ge
Si
x d x
of 22 45
Equation 6.3
Equation 6.4
If in the equation (The accurate harmonic mean form):
Equation 6.5
We make the approximation
Equation 6.6
Then the Stokes form of the equation becomes
Equation 6.7
Thus we see for this approximation there are two integrals as well:
Equation 6.8
Equation 6.9
∫
1
0
∫
1
0
Si
B
(As − G a)d yd z ≈
1
3
1
(Ge − Si )
∫
Ge
Si
x d x
∫
1
0
∫
1
0
Ge
B
(P − Al )d x dz ≈
2
3
1
(Ge − Si )
∫
Ge
Si
yd y
Si
B
(As − G a) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
2SiGe
Si + Ge
≈ G e − Si
∫
1
0
∫
1
0
[
Si
B
(As − G a) +
Ge
B
(P − Al )
]
d yd z =
∫
Ge
Si
d x
∫
1
0
∫
1
0
Si
B
(As − G a)d yd z =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al )d ydz =
2
3
∫
Ge
Si
dz
of 23 45
We pull these AI elements out of the periodic table of the elements to make an AI periodic
table:!
We now notice we can make a 3 by 3 matrix of it, which lends itself to to the curl of a vector
field by including biological elements carbon C (above Si):!
=!
=!
!
Which resulted in Stokes theorem (Beardsley, Essays In Cosmic Archaeology Volume 3):!
!
Where!
!
!
!
We were then able to write this with product notation!
!
i
j
k
∂
∂x
∂
∂y
∂
∂z
(−C ⋅ P)y (Si ⋅ Ga)z (Ge ⋅ As)y
(Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
[
(72.64)(74.92) − (28.09)(69.72)
]
i +
[
(12.01)(30.97)
]
k
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a = exp
(
1
Ge − Si
∫
Ge
Si
ln(x)d x
)
∇ ×
u = (Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
d
a =
(
zd yd z
i + ydydz
k
)
u = (−C ⋅ P)y
i + (Si ⋅ Ge)z
j + (G a ⋅ As)y
k
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a =
n
n
∏
i=1
x
i
of 24 45
While we have the AI BioMatrix!
Which we used to formulate a similar equation (Beardsley, Essays In Cosmic Archaeology
Volume 2). We can form another 3X3 matrix we will call the electronics matrix (Beardsley,
Cosmic Archaeology, Volume Three):!
!
We can remove the 5th root sign in the above equation by noticing!
!
=(28.085)(72.64)(12.085)(107.8682)(196.9657)=!
!
Where we have substituted carbon (C=12.01) the core biological element for copper (Cu).!
But since we have:!
!
We take the ratio and have!
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ Cu ⋅ Ag ⋅ Au
523,818,646.5
g
5
mol
5
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u ) ⋅ d
a = 170,535,359.662( g /mol )
5
of 25 45
!
Almost exactly 3 which is the ratio of the perimeter of regular hexagon to its diameter used to
estimate pi in ancient times by inscribing it in a circle:!
!
Perimeter=6!
Diameter=2!
6/2=3!
!
Thus we have the following equation…!
Equation 6.10 !
Showing the calculation using the most accurate data possible…!
Ge=72.64!
As=74.9216!
Si=28.085!
Ga=69.723!
C=12.011!
P=30.97376200!
=!
=!
523,818,646.5
170,535,359.662
= 3.0716
π = 3.141...
π
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u ) ⋅ d
a =
5
∏
i=1
x
i
(Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
[
(72.64)(74.9216) − (28.085)(69.723)
]
i +
[
(12.011)(30.97376200)
]
k
of 26 45
!
!
!
!
!
=154,082,837.980+16,452,521.6822=!
!
=!
(28.085)(72.64)(12.085)(107.8682)(196.9657)=!
!
Where we have substituted carbon C=12.01 for copper Cu. We use Cu, Ag, Au because they
are the middle column of our electronics matrix, they are the finest conductors used for
electrical wire. We use C, Si, Ge because they are the middle column of our AI Biomatrix. Si
and Ge are the primary semiconductor elements used in transistor technology (Artificial
Intelligence) and C is the core element of biological life. We have!
!
!
Perimeter/Diameter of regular hexagon = 3.00!
3,484.134569
(
g
mol
)
2
i + 372.025855
(
g
mol
)
2
k
∫
Ge
Si
∫
Ge
Si
(
3,484.134569
(
g
mol
)
2
i + 372.025855
(
g
mol
)
2
k
)
⋅
(
zd yd z
i + ydydz
k
)
∫
Ge
Si
∫
Ge
Si
(
3,484.134569
(
g
mol
)
2
⋅ zdzdy + 372.025855
(
g
mol
)
2
⋅ ydzd y
)
∫
Ge
Si
3,484.134569
(
(72.64 − 28.085)
2
2
)
dy +
∫
Ge
Si
372.025855y ⋅ (72.64 − 28.085)d y
3458261.42924
(
g
mol
)
4
(72.64 − 28.085) + 16575.6119695
(
g
mol
)
3
(
(72.64 − 28.085)
2
2
)
170,535,359.662
(
g
mol
)
5
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ C ⋅ Ag ⋅ Au
523,818,646.5
g
5
mol
5
523,818,646.5
170,535,359.662
= 3.0716
π = 3.141...
of 27 45
!
The same value as our 3.0716 if taken at two places after the decimal.!
Thus we are interested in:!
Equation 6.11 !
Equation 6.12 #
3.141 + 3.00
2
= 3.0705
π
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u ) ⋅ d
a =
5
∏
i=1
x
i
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ C ⋅ Ag ⋅ Au
of 28 45
Section 7.0 Natural Abundances
We now formulate what I call Giordano’s Relationship: Warren Giordano writes in his paper The
Fine Structure Constant And The Gravitational Constant: Keys To The Substance Of The Fabric
Of Space, March 21, 2019:!
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where ‘ ’
is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational constant
numerically, but neglecting any units.
Let’s do that!
(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js!
And it works, G is:!
G=6.67408E-11 N(m2/kg2)!
This is like saying gravitational constant masses equal ten to the 23 Planck velocities:!
!
!
Where m1 is a kilogram and v1 is 1 meter per second. Avogadro’s number is!
!
But, if we are working with kilograms we say!
!
But H=0.001 kg/mol not 1.00 g/mol and carbon is C=0.012 kg/mol not 12.01 g/mol!
Thus if H is the molar mass of hydrogen (1 g/mol=0.001kg/mol) then there are 6E26 atoms in a
kilogram of it. We have!
!
Remembering we now say in this system of units if we are working in
kg.!
But can we say!
!
And our equations become!
h
1 + α
α
10
23
Gm
1
= 10
23
h(1 + α)(v
1
)
Gm
1
= 10
23
h(1 + α)(v
1
)
6E 23atom s /m ol = N
A
N
A
= 6E26atom s/m ole
6 ⋅ secon ds
kilograms
meter
= 1000h
N
A
G
(1 + α)
N
A
= 6E26atom s/m ol
N
A
H
= 6E23
of 29 45
!
!
Where E is any element and N is some number. Let us reformulate this as:!
Equation 7.1 !
Where and H=1 gram/atom!
Because for hydrogen 1 proton is molar mass 1 gram, for carbon 6 protons is 6 grams and so
on for 6E23 atoms per gram. Thus,…!
!
Since grams and atom cancel we can work in grams even though our equations are in
kilograms. Let us not write H, since formally it is grams per mole of hydrogen but write !
!
We have:!
!
Or,…!
Equation 7.2 !
Where!
!
Let us say we were to consider Any Element say carbon . Then in general!
Equation 7.3 !
We have!
6 ⋅ secon ds
kilograms
2
meter
= h
N
A
H
(1 + α)
G
N ⋅ k ilogram s
2
secon d s
meter
= h
N
A
E
(1 + α)
G
h
(1 + α)
G
N
A
H = 6.0003
kg
2
⋅ s
m
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
N
A
H = 6.02E 23
atom s
gra m
⋅
1gra m
atom
= 6.02E 23
ℍ = 1
gra m
atom
h
(1 + α)
G
⋅ N
A
ℍ = 6.0003kg
2
⋅
s
m
h(1 + α)N
A
ℍ = 6Gx
x = 1.00kg
2
s
m
𝔼
ℂ
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
of 30 45
and !
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have 12-6=6 grams of protons in the 12 grams of protons and neutrons.
Thus!
!
And it follows that!
!
We see in general since the atomic number Z is the number of protons in an atom that in
general this holds for all elements because!
!
And,!
!
Therefore we always have:!
Equation 7.4 !
Let us consider then equation 5. We multiply both sides by .!
!
That is we have!
Equation 7.5 !
We will return to this answer, but me we look at the significance of time in Quantum Mechanics
first.!
Thus returning to theorem 1.0 and considering equation 2.1 And 7.3!
ℂ =
6gra m s
6proton s
N
A
=
6(6E 23protons)
6gra m s
N
A
ℂ = 6E 23
h
(1 + α)
G
⋅ N
A
ℂ = 6.0003kg
2
⋅
s
m
𝔼
N
A
=
Z ⋅ 6E 23proton s
Z ⋅ gram s
𝔼 =
Z ⋅ gram s
Z ⋅ protons
N
A
⋅ 𝔼 = 6E 23
r
p
m
2
p
r
p
m
2
p
= 6.000kg
2
s
m
⋅
0.833E − 15m
(1.67262E − 27kg)
2
= 1.7865E39s
r
p
m
3
p
⋅ h
(1 + α)
Gx
⋅ N
A
𝔼 = 1.7865E 39s
of 31 45
Theorem 1.0 The Periodic Table of the Elements employs six fold symmetry. The periods of the
periodic table are periodic over 18 groups, which is 3 times 6, the three being the components
of , the number of values needed to specific a point in such a space.!
!
!
Thus in 2.1 we have 6 protons and in 7.3 we have . Equation 1 is determined by
. !
Buckminster Fuller writes in Synergetics:!
205.01 The geometrical model of energy configurations in synergetics is developed from a
symmetrical cluster of spheres, in which each sphere is a model of a field of energy all of whose
forces tend to coordinate themselves shuntingly or pulsatively, and only momentarily in
positive or negative asymmetrical patterns relative to, but never congruent with, the eternality of
the vector equilibrium…This closest packing of spheres in 60-degree angular relationships
demonstrates a finite system in Universal geometry.
ℝ
3
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6proton s
h
(1 + α)
G
N
A
H = 6.000
kg
2
⋅ s
m
6
kg
2
⋅ s
m
t
1
= 1s
of 32 45
#
of 33 45
Thus we see carbon is 8 tetrahedra and 6 octahedra, each of which is 2 tetrahedra I found
them precisely this in their two dimensional analogs the regular dodecahedron and regular
octagon:!
We see that the atomic radius of silicon the core element of artificial intelligence (transistor
technology) fits together with the core element of biological life carbon if the silicon is taken as
inscribed in a regular dodecagon and the carbon is taken as inscribed in a regular octagon. We
have:!
, , !
, , !
Apothem: !
For a regular dodecagon:!
!
The radius of a silicon atom is Si=0.118nm and that of carbon is
C=0.077nm:!
!
!
!
This has an accuracy %!
But, we still have the question of 1 second in equation 2.1:!
!
We see that the vector equilibrium employs 60-degree coordination. And that it involves a layer
of 12 spheres. However we have a 12 hour daytime and and 12 hour night time, which is a 24
hour day (Earth rotates through 15 deg an hour, dodecagon). But there are 60 minutes in an
hour and 60-degree coordination is connected to that in that the earth rotates through 360
degrees in 24 hours and 360/60 degrees is our six-fold symmetry of the periodic table. Each
hour then of the 24 hours is divided by 60 minutes and each minute divided by 60 is 1 second.
D = 1 + 2x
x
2
+ x
2
= 1
2
2x
2
= 1
2x
2
= 1
x = 2 /2
D = 1 + 2
a = (1 + 2)/2 = 1.2071
a =
s/2
ta n(θ /2)
=
0.5
ta n(15
∘
)
= 1.866
Si
C
=
0.118
0.077
= 1.532
a
S
i
a
C
=
1.866
1.2071
= 1.54585
1.532
1.54585
100 = 99
1
t
1
α
2
m
p
h4πr
2
p
Gc
= 6proton s
of 34 45
We have our one second. Indeed I find the size of the solar system is connected to Silicon
through its counterpart as a semiconductor material germanium. Let’s look at that:!
While we have!
!
Is a number of protons!
!
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided
by the mass of a proton. But these masses can be considered to cancel and leave pure
number. We have that!
!
!
Interestingly 78% is the percent of N2 in the atmosphere and 21% is the percent of O2 in the
atmosphere (by volume). These are the primary constituents that make it up. The rest is
primarily argon and CO2. This gives the molar mass of air as a mixture is:!
!
Now interestingly, I have found this connected to the solar system by considering a mixture of
silicon, the primary constituent of the Earth crust, and germanium just below it in the periodic
table, in the same proportions of 78% and 21%. Silicon is also the primary second generation
semiconductor material (what we use today) and germanium is the primary first generation
semiconductor material (what we used first). The silicon is directly below our carbon of one
proton-second, silicon directly below that, and germanium directly below that, in the periodic
table. So we are moving directly down the periodic table in group 14. The density of silicon is
2.33 g/cm3 and that of germanium is 5.323 g/cm3. Let us weight these densities with our 0.21
and 0.78:!
!
Now consider this the starting point for density of a thin disc decreasing linearly from the Sun
to Pluto (49.5AU=7.4E14cm). Thus,…!
!
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
1
α
2
m
p
h4πr
2
p
Gc
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
0.78N 2 + 0.21O2 = 29.0g /m ol
0.21Si + 0.78Ge = 4.64124g/cm
3
ρ(r) = ρ
0
(
1 −
r
R
)
of 35 45
Thus,…!
!
Or,…!
!
Thus,…!
!
If we add up the masses of the planets in our solar system they are 2.668E30 grams.!
Since!
!
Meaning mixing germanium and silicon in the same proportion that occurs with N2 and O2 in
the atmosphere and with!
!
!
Where!
!
In the first integral. See the following pages to see the computation of the mass of the planets
in the solar system…!
!
As we can see Jupiter carries the majority of the mass, Saturn a pretty large piece, and
somewhat large Uranus and Neptune. We don’t even list Pluto’s mass. When we consider!
M =
∫
2π
0
∫
R
0
ρ
0
(
1 −
r
R
)
rdrdθ
M =
πρ
0
R
2
3
M =
π(4.64124)(7.4E14)
2
3
= 2.661E30
2.661
2.668
(100) = 99.736
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
6 =
1
α
2
m
p
h4πr
2
p
Gc
of 36 45
!
!
We are considering and . These come from!
!
!
From 30 degrees to 45 degrees is 15 degrees. The Earth rotates through 360/24 is 15 degrees
per hour. The hour is divided into 60 minutes and minutes into 60 seconds…!
We have said!
!
For a thin disc. Thus we have a definition for the radius of the solar system, :!
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
3
2
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
−
π − 4
2 2
= 0.21
2
3
2 cos(45
∘
) = 2
2 cos(30
∘
) = 3
M =
πρ
0
R
2
3
R
s
of 37 45
Equation 7.6 !
Where!
Equation 7.7 !
Equation 7.8 !
Equation 7.9 !
Equation 7.10 !
is the mass of all the planets. We also have that the molar mass of air to the molar mass of
water is approximately the golden ratio. The interesting thing is we determine a definition for
the radius of the solar system and predict the hydrocarbons (backbones of life) all in one fell
swoop. Thus since if the universe is mostly hydrogen and helium if we can predict their relative
abundances with proton-seconds then we can feel quite certain we are formulating things
correctly, consider the following gaussian distribution illustrated…#
R
s
=
3M
p
π(0.78Ge + 0.21Si )
1
α
2
m
p
h4πr
2
p
Gc
∫
3
2
1
t
2
dt = 0.78
∫
3
2
cos
−1
(x /2)d x = 0.21
air = 0.78N
2
+ 0.21O
2
air
H
2
O
≈ Φ
M
p
of 38 45
#
of 39 45
We consider a Gaussian wave-packet at t=0:!
!
We say that d which in quantum mechanics would be the delocalization length when squared is
. A is the amplitude and we might say it is . We write the wave packet as a
Fourier transform which is:!
!
We use the identity that gives the integral of a quadratic:!
!
Solve the equation!
!
With the initial condition!
!
A plane wave is the solution:!
!
Where, !
The wave-packet evolves with time as!
!
Calculate the Gaussian integral of dp!
and !
The solution is:!
ψ (x,0) = Ae
−
x
2
2d
2
Si − C
C
Ga − P
C
ψ (x,0) = Ae
−
x
2
2d
2
=
∫
dp
2π ℏ
ϕ
p
e
i
h
px
∫
∞
−∞
e
−α
2
x+βx
d x =
π
α
e
β
2
4α
iℏ∂
t
ψ (x, t) =
p
2m
ψ (x, t)
ψ (x,0) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
px
e
i
ℏ
( px−ϵ( p)t)
ϵ( p) =
p
2
2m
ψ (x, t) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
( px−
p
2
2m
t)
α =
d
2
2ℏ
2
+
it
2mℏ
β =
i x
ℏ
of 40 45
where !
We notice here one of the things you can do with our equation for proton-seconds is integrate
from 0.5 sec to 1 sec and you get one which multiplied by the constant which is six yields six.
Now look up 0.5 seconds from the data output from the program and it is magnesium, then go
to one second and it is carbon, thus the integral from magnesium in time to carbon in time is
carbon in protons. Now consider life as we know it is based on carbon because it has four
valence electrons, but it is not based on silicon, which has four valence electrons as well,
because in the presence of oxygen it readily forms SiO2 (sand or glass) leaving it unavailable to
nitrogen, phosphorus, and hydrogen to make make amino acids the building blocks of life. But
silicon can be doped with phosphorus, boron, gallium, and arsenic to make semiconductors --
transistor technology from which we can build artificial life (artificial intelligence, AI). We can
integrate over many time ranges to explore millions of more facets to the equation:!
!
!
!
We should say!
or !
The way I am using equation 1 is . We have:!
Equation 3.14!
!
ψ
2
= exp
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
τ =
m d
2
ℏ
1
α
2
m
p
h4πr
2
p
Gc
∫
t
C
t
Mg
1
t
2
dt = 6 = carbon(C )
t
Mg
= 0.5secon ds
t
C
= 1secon d
d =
Si − C
C
=
4
3
d
2
=
16
9
τ = d
2
ψ
2
= exp −
C
2
x
2
(Si − C )
2
⋅
1
1 +
[
ℏC
2
m(Si − C)
2
]
2
t
2
of 41 45
Thus for hydrogen:!
!
=74%.!
For Helium:!
=26%!
This is interesting because the Universe is about 74% Hydrogen and 24% Helium, the rest of
the elements making up the other 2%. Thus we can say or . We have
multiplied the first by 1 for Hydrogen element 1 (1 proton), and the second by 2 for helium
element 2 (2 protons). In a sense then, the probabilities represent the probability of finding
hydrogen and helium in the Universe. Hydrogen much of the helium were made theoretically in
the Big Bang of the big bang theory. The other elements were synthesized from these by the
stars. The difference between hydrogen and helium is that some of the helium was made in
stars, that may be why it is 26% not 24% because it could include the remaining 2% of
elements in the universe. It must be kept in mind the data we have on the universe as a whole
for relative abundances can only currently be ballpark figures.#
ψ
2
= exp −
9
16
x
2
⋅
1
1 +
ℏ
2
81
m
2
256
t
2
ψ
2
= (1)exp −
9
16
(1proton)
2
⋅
1
1 +
(0.075)81
(1)256
(6secon d s)
2
ψ
2
= (2)exp −
9
16
(2proton)
2
⋅
1
1 +
(0.075)81
(4)256
(3secon d s)
2
ℏ
2
= 0.075
ℏ = 0.27386
of 42 45
Section 8.0 Formulation of Proton-Seconds
We can actually formulate this differently than we have. We had!
!
!
But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2
are lower and upper limits in an integral, then we have:!
Equation 8.1 !
This Equation is the generalized equation we can use for solving problems.!
Essentially we can rigorously formulate the notion of proton-seconds by considering!
Equation 8.2 !
Is protons-seconds squared where current density is and ( can also be
). We say!
Equation 8.3. !
Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a
number of protons. It is!
Equation 8.4 !
Dividing Equation 4.2 through by by t:!
Equation 8.5 !
!
Which is proton-seconds. Dividing through by t again:!
1
t
1
1
α
2
m
p
h4πr
2
p
Gc
= 6proton s
1
t
6
1
α
2
m
p
h4πr
2
p
Gc
= 1proton
1
α
2
m
p
h4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
∫
t
qdt = t
2
∬
S
ρ(x, y, z)d xdy
J = ρ
v
ρ = Q /m
3
ρ
Q /m
2
Q =
∫
V
ρdV
ℕ =
1
α
2
m
p
h4πr
2
p
Gc
1
α
2
m
p
h4πr
2
p
Gc
∫
t
dt
t
= t
∬
S
ρ(x, y, z)d xdy
of 43 45
Equation 8.6 !
We see that if where and then J is I/m2 (current per square
meter) is analogous to amperes per per square meter which are coulombs per second through
a surface. Thus we are looking at a number of protons per second through a surface. Thus we
write:!
!
Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to
protons in the output from our program and 1.0 seconds is carbon (C). We see we have the
following theorem:!
Equation 8.7 !
So as an example,…!
!
Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second
which is carbon and we have protons per square meter. Charge conservation gives:!
!
Is 9-9=0. In general…!
!
!
#
1
α
2
m
p
h4πr
2
p
Gc
∫
t
dt
t
2
= proton s
J = ρ
v
ρ = Q /m
3
v = m /s
1
α
2
m
p
h4πr
2
p
Gc
∫
t
C
t
Mg
dt
t
2
= 6
∫
1.0
0.5
t
−2
dt = − 6(1 − 2) = 6
1
α
2
m
p
h4πr
2
p
Gc
∫
t
dt
t
3
=
∬
S
J ⋅ d
S
1
α
2
m
p
h4πr
2
p
Gc
∫
1.0
0.5
dt
t
3
=
∬
S
J ⋅ d
S = − 3
(
1 −
1
0.25
)
= 9
proton s
secon d
∂ρ
∂ t
+ ∇ ⋅
J = 0
J (x, y, z) = (0,0,J ) = − J
k
d
S = d x d y
k
J ⋅ d
S = (0,0,J ) ⋅ (0,0,d xdy) = − Jd xdy
of 44 45
!
We consider the surface through which J passes a circle of radius R where J decreases linearly
as r with at the center we have:!
Equation 8.8 !
!
!
J
0
∬
S
J ⋅ d
S = J
0
∫
2π
0
∫
R
0
(
1 −
r
R
)
rdrdθ =
π J
0
R
2
3
of 45 45
The Author
